Question 283657
If a gambler rolls two dice and gets a sum of 7, he wins $10, and if he gets a sum of 4, he wins $20. The cost to play the game is $5. What is the expectation (to the nearest cent) of this game?
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There are three possible outcomes of playing the game:

A. He can pay $5 to play, roll a sum of 7, win $10, and net $5.

B. He can pay $5 to play, roll a sum of 4, win $20, and net $15.

C. He can pay $5 to play, roll a sum of neither 4 nor 7, and net -$5 (a loss).

Now we need to calculate the probabilities of doing those.

Here is the sample space of rolling a pair of dice:


(1,1)   (1,2)   <font color = "blue">(1,3)</font>   (1,4)   (1,5)   <font color = "red">(1,6)</font> 


(2,1)   <font color = "blue">(2,2)</font>   (2,3)   (2,4)   <font color = "red">(2,5)</font>   (2,6)


<font color = "blue">(3,1)</font>   (3,2)   (3,3)   <font color = "red">(3,4)</font>   (3,5)   (3,6)


(4,1)   (4,2)   <font color = "red">(4,3)</font>   (4,4)   (4,5)   (4,6)


(5,1)   <font color = "red">(5,2)</font>   (5,3)   (5,4)   (5,5)   (5,6)


<font color = "red">(6,1)</font>   (6,2)   (6,3)   (6,4)   (6,5)   (6,6)

There are 36 posasible rolls.  The 6 red ones have sum 7, the 3 blue ones
have sum 4, the 27 black ones have some other sum.  So 

The probability of rolling a 7 is {{{6/36}}} or {{{1/6}}}. P(A) = {{{1/6}}}

The probability of rolling a 4 is {{{3/36}}} or {{{1/12}}}. P(B) = {{{1/12}}}

The probability of rolling neither of those sum is {{{27/36}}} or {{{3/4}}}
   P(C) = {{{3/4}}}

To find the expectation, multiply each win by the probability of winning it,
then add them.

{{{Expectation = ("$10")(1/6) + ("$20")(1/12) + ("-$5")(3/4)}}}

{{{Expectation = 10/6 + 20/12 - 15/4}}}

{{{Expectation = 20/12 + 20/12 - 45/12}}}

{{{Expectation = -5/12=-.41}}}{{{2/3}}}

That means if one plays the game many times he will have
averaged losing {{{41&2/3}}} cents per game.

Edwin</pre>